áp dụng BĐT : ab \(\le\dfrac{a^2+b^2}{2}\). dấu "=" xảy ra khi a = b. ta có :
xz+yz=z(x+y) \(\le\dfrac{z^2+\left(x+y\right)^2}{2}=\dfrac{x^2+2xy+y^2+z^2}{2}\le xy+1\)
dấu "=" xảy ra \(\Leftrightarrow\)\(\left\{{}\begin{matrix}x^2+y^2+z^2=2\\z=x+y\end{matrix}\right.\)
ta có |xy| \(\le\dfrac{x^2+y^2}{2}< \dfrac{x^2+y^2+z^2}{2}\le1\)
nếu xy\(\ge\)-1 \(\Leftrightarrow\)\(\left\{{}\begin{matrix}\left|x\right|=\left|y\right|\\xy=-1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=1;y=-1\\x=-1;y=1\end{matrix}\right.\)
\(\Rightarrow\) S= 2016xy - (yz+xz) \(\ge\) 2016xy - xy - 1 \(\Rightarrow\) 2015xy - 1 \(\Rightarrow\)S\(\ge\)-2016
vậy Min S = -2016 \(\Rightarrow\)\(\left\{{}\begin{matrix}x=1;y=-1;z=0\\x=-1;y=1;z=0\end{matrix}\right.\)
